1. 2.0 Statics of Particles
Dr. Engin Aktaş

2. Forces on a Particle 50 N A Line of Action magnitude
Point of Application
50 N A
30o
direction
Since the vector has a well defined point of application, it is a fixed vector, therefore can not be moved without modifications
Dr. Engin Aktaş

3. Properties of Vector Addition
Commutative
P+Q=Q+P P Q P A
P+Q Q A
Q+P P A Q
Associative Q Q P P S S
Q+S
P+Q A A
P+Q+S
P+Q+S
P+Q+S=(P+Q)+S=P+(Q+S)
Dr. Engin Aktaş

4. Resultant of several concurrent forcesP
Using polygon rule P A Q Q A S S
P+Q+S
Dr. Engin Aktaş

5. Resolution of a force into componentsQ Q F F A A P P F Q A P
Dr. Engin Aktaş

6. Example (Beer & Johnston)
Q=60 N
The two forces P and Q act on a bolt A. Determine their resultant.
25o
P=40 N A
20o A P Q R a
180-25=155o
25o B
20o
Law of cosines
R2=P2+Q2-2PQcosB
R2=(40N)2+(60N)2-2(40N)(60N)cos1550
R=97.73 N
Law of sines
Dr. Engin Aktaş

7. Example (cntd.) The answer is a=35.0o R=97.7 kN Law of sines A=15.04o
a=20o+A=35.04o
The answer is
R=97.7 kN
a=35.0o
Dr. Engin Aktaş

8. Example (Beer and Johnston)C 1 2
30o
45o
A barge is pulled by two tugboats. If the resultant of the forces exerted by tugboats is a 25 kN force directed along the axis of barge, determine the tension in each of the ropes.
Using law of sines
R=25 kN
30o
45o
105o T2 T1
Dr. Engin Aktaş

9. Rectangular Componentsy F Fy q x Fx
Dr. Engin Aktaş

10. y Fy F q Fx x
Dr. Engin Aktaş

11. Unit Vectors F Fy=Fyj j Fx=Fxi i y Magnitude=1 x 01.04.2017
Dr. Engin Aktaş

12. Example (Beer and Johnston)
F=(3.5 kN)i+(7.5kN)j
Determine the magnitude of the force and angle q. q A y
Fy=7.5 kN F q x A
Fx=3.5 kN
Dr. Engin Aktaş

13. Addition of Forces by Summing x and y components
F2=80 N
F1=150 N
Determine the resultant of the forces on the bolt
F2 cos 20o
-(F2 sin 20o)
30o A
F1 sin 30o
F1 cos 30o x
15o
F4=100 N
F4 cos 15o
-(F4 sin 15o)
F3=110 N
-F3
Forces
Magnitude, N
x Component, N
y Component, N F1
150
+129.9
+75.0 F2 80
-27.4
+75.2 F3
110
-110.0 F4
100
+96.6
-25.9
Rx=199.1
Ry=+14.3
Dr. Engin Aktaş

14. Example (cntd) R=Rxi+Ryj R=(199.1N)i+(14.30N)j R a Ry=14.30 N
Rx=199.1 N
Dr. Engin Aktaş

15. Equilibrium of a Particle
When the resultant of all the forces acting on a particle is zero, the particle is in equilibrium. x y
F1=300 N
F4=400 N
30o
F2=173.2 N
F4=400 N A
F1=300 N
F2=173.2 N
F3=200 N
F3=200 N
R=SF=0
(SFx)i+ (SFy)j=0
SFx=0
SFy=0
SFx=300 N -(200 N) sin30o-(400 N) sin 30o=300 N -100 N -200 N = 0
SFy= N -(200 N) cos30o+ (400 N) cos 30o= N N N = 0
Dr. Engin Aktaş

16. Free-Body Diagram
A sketch showing the physical conditions of the problem is known as space diagram. A B C
50o
30o
Space Diagram
Choose a significant particle and draw a separate diagram showing that particle and forces on it. This is the Free-Body diagram.
TAC
30o
50o
TAB
736 N
Using law of sines
TAB
40o
736 N
80o
60o
TAB=647 N
TAC
Free-Body Diagram
TAC=480 N
Dr. Engin Aktaş

17. Analytical Solution Two unknowns TAB and TAC Equilibrium Equations
SFx=0 and SFy=0
The system of equations then
Solving the above system (2 unknowns 2 equations)
TAB=647 N
TAC=480 N
Dr. Engin Aktaş

18. Sumary of Solution techniques
equilibrium under three forces
may use force triangle rule
equilibrium under more than three forces
may use force polygon rule
If analytical solution is desired
may use equations of equilibrium
Dr. Engin Aktaş

19. Example (Beer and Johnston)
As a part of the design of a new sailboat, it is desired to determine the drag force which may be expected at a given speed. To do so, a model of the proposed hull is placed in a test channel and three cables are used to keep its bow on the centerline of the channel. Dynamometer reading indicate that for a given speed, the tension is 200 N in cable AB and 300 N in cable AE. Determine the drag force exerted on the hull and the tension in cable AC. B C A
60o
20o
Flow E
Equilibrium condition
Start with drawing Free-Body Diagram
Resultant of the all forces should be zero
R=TAB+TAC+TAE+FD=0
20o
TAC
Let’s write all the forces in x and y components
TAB=200 N
60o
TAB=-(200 N) sin 60o i+(200 N) cos 60o j FD
TAB= i+100 j
TAC= (TAC) sin 20o i+(TAC) cos 20o j
Unknowns; FD, TAC
TAC= 0.342(TAC) i (TAC) j
TAE=-(300 N) j
TAE=300 N
FD=FD i
Dr. Engin Aktaş

20. Example (cntd.) R=TAB+TAC+TAE+FD=0
R=( N)i N j (TAC) i (TAC) j + (-300 N) j + FD i =0
R={ (TAC) + FD} i + { (TAC) + (-300 N)} j = 0
SFx=0
N (TAC) + FD = 0
(TAC) + (-300 N) = 0
SFy=0
TAC=213 N
FD=100 N
Dr. Engin Aktaş

21. Forces in Space V j k Vzk qy Vyj qz qx Vxi i V=Vxi+Vyj+Vzk Vx=V cos qx
Vy=V cos qy
Vz=V cos qz
Dr. Engin Aktaş

22. Direction cosines l = cos qx m = cos qy n = cos qz Vz=nV Vy=mV Vx=l V
V 2=Vx2+Vy2+Vz2
l 2 + m 2 + n 2=1
Dr. Engin Aktaş

23. B(xB, yB, zB) V n A(xA, yA, zA) V= (xB-xA) i (yB-yA) + j + (zB-zA) k
Unit vector along AB
Dr. Engin Aktaş

24. Rectangular Coordinates in Spacey y B B A Fy qy Fy F Fx O x O x f D f Fz Fh Fh C E C
Fy=F cosqy
Fx = Fh cosf = F sinqy cosf z z
Fh=F sinqy
Fz = Fh sinf = F sinqy sinf
F 2 = Fy2 + Fh2
Fh2 = Fx2 + Fz2
Dr. Engin Aktaş

25. Problems
Dr. Engin Aktaş

26. Problem 1-(Meriam and Kraige)y
F=1800 N 4
The 1800 N force F is applied at the end of the I-beam. Express F as a vector using the unit vectors i and j. A 3 x
First let’s find the x and y components of the force F. z
Fx= N 3/5 = N
Fy= N 4/5 = N
F = Fx i+ Fy j = i j N
Dr. Engin Aktaş

27. Problem 2-(Meriam and Kraige)
Air flow D L C
The ratio of the lift force L to the drag force D for the simple airfoil is L/D = 10. If the lift force on a short section of the airfoil is 50 N, compute the magnitude of the resultant force R and the angle q which it makes with the horizontal.
L= 50 N
tan q = 50/5 q C
q = tan-110 =84.3o
D=5 N
F = (502+52)0.50 = 50.2 N
Dr. Engin Aktaş

28. Problem 3-(Meriam and Kraige)y
The gusset plate is subjected to the two forces shown. Replace them by two equivalent force, Fx in the x-direction and Fa in the a direction. Determine the magnitudes of Fx and Fa. A x
10o
From the law of cosines
800 N
900 N a
R2= (800)(900) cos75
25o
45o
R=1040 N A
From the law of sines
800 N
10o
10o a
66.7o
45o
113.3o A Fx Fa
21.7o
1040 N R
65o
25o
900 N
Dr. Engin Aktaş

29. Problem 4-(Meriam and Kraige)C
50 m
40 m
20 m
The guy cables AB and AC are attached to the top of the transmission tower. The tension in cable AC is 8 kN. Determine the required tension T in cable AB such that the net effect of the two cable tensions is a downward force at point A. Determine the magnitude R of this downward force
This time let’s use another approach. Since the resultant force is downward the sum of horizontal components of the two forces should add up to zero.
8 kN R
The magnitude of R
Dr. Engin Aktaş

30. Problem 5-(Beer and Johnston)
500 mm
1375 mm
Two cables are tied together at C and loaded as shown. Determine the tension in AC and BC. A B
1200 mm
Let’s draw the Free Body Diagram (FBD) at C C y
1600 kg
TBC
TAC x C
W=1600*9.81=15700N
Writing Equilibrium Equations
SFx=0
SFy=0
TAC=12470 N
TCB=6370 N
Dr. Engin Aktaş

31. Problem 6-(Beer and Johnston)C
5.5 m
8 m
4 m
13.5 m y
A precast-concrete wall section is temporarily held by the cables shown. Knowing that tension is 4200 N in cable AB and 6000 N in cable AC, determine the magnitude and direction of the resultant of the forces exerted by cables AB and AC on stake A. x
Coordinates of points A, B and C
A (8, -4, -5.5)
B (0, 0, 0)
C (0, 0, -13.5)
AB=(0-8) i + (0-(-4)) j + (0-(-5.5)) k = -8 i + 4 j k z
AC=(0-8) i + (0-(-4)) j + (-13.5-(-5.5)) k = -8 i + 4 j - 8 k
Dr. Engin Aktaş

32. Problem 6-(Beer and Johnston)C
5.5 m
8 m
4 m
13.5 m y
A precast-concrete wall section is temporarily held by the cables shown. Knowing that tension is 4200 N in cable AB and 6000 N in cable AC, determine the magnitude and direction of the resultant of the forces exerted by cables AB and AC on stake A. x z
R = FAB + FAC
R =( ) i +( ) j + ( ) k = i j k
R = 8250 N x y z
Direction cosines
l = 0.873
m = 0.436
n = 0.218
Dr. Engin Aktaş

33. Problem 7-(Beer and Johnston)y x z B A O C D
700 mm
600 mm
1125 mm
650 mm
450 mm
Let’s draw FBD first
FAC
FAD
FAB W
Unknowns : FAD, FAC and W
Coordinates of points A, B, C and D
A (0, -1125, 0)
B (700, 0, 0)
C (0, 0, -600)
D (-650, 0,450)
A crate is supported by three cables as shown. Determine the weight W of the crate knowing that the tension in cable AB is 4620 N.
AB=((700-0)i+(0-(-1125))j+0k=700i+1125j+0k
AB=( )0.5=1325
nAB=(700i+1125j+0k)/1325=0.5283i j
FAB=4620(0.5283i j) N
Dr. Engin Aktaş

34. Problem 7-(Beer and Johnston)y x z B A O C D
700 mm
600 mm
1125 mm
650 mm
450 mm
Coordinates of points A, B, C and D
A (0, -1125, 0)
B (700, 0, 0)
C (0, 0, -600)
D (-650, 0,450)
AC=(0i+(0-(-1125))j+(-600-0)k=0i+1125j-600k
AC=( )0.5=1275
nAC=(0i+1125j-600k)/1275=0.8824j k
FAC=FAC(0.8824j k) N
AD=((-650-0)i+(0-(-1125))j+(450-0)k=-650i+1125j+450k
AD=( )0.5=1375
nAD=(-650i+1125j+450k)/1375= i j k
FAD=FAD( i j k) N
W=-W j
Dr. Engin Aktaş

35. Problem 7-(Beer and Johnston)y x z B A O C D
700 mm
600 mm
1125 mm
650 mm
450 mm
FAB+FAC+FAD+W=0
4620(0.5283i j) +FAC(0.8824j k)+FAD( i j k)-W j=0
FAD=0
FAD= 5160 N
FAC *5160=0
FAC= 3590 N
W= N
W = 0
Dr. Engin Aktaş

36. Problem 8-(Beer and Johnston)y D
600 mm C
200 mm A x
200 mm B
200 mm
400 mm z
A 16 kg triangular plate is supported by three wires as shown. Determine the tension in each wire .
Dr. Engin Aktaş